Respuesta :
Assume that air resistance may be ignored.
All measurements are positive upward.
g = acceleration due to gravity.
Case A; The ball is thrown vertically upward.
The time, t₁, to reach maximum height is one half of the time of flight.
Because the vertical velocity is zero at maximum height,
V₀ - gt₁ = 0
t₁ = V₀/g.
The time of flight is
t = 2t₁ = (2V₀)/g. (1)
Let v = the vertical velocity with which the ball strikes the ground. Then
v² = V₀² + 2(-g)(-h)
v = √(V₀² + 2gh) (2)
Case B: The ball is thrown vertically downward.
The time of flight, t, is given by
-h = -V₀t + (-g)t²
gt² + V₀t - h = 0
t = 1/(2g)[-V₀ +/- √(V₀² + 4gh)]
Reject negative time
[tex]t=- \frac{V_{0}}{2g} + \frac{V_{0}}{2g} \sqrt{1+ \frac{4gh}{V_{0}^{2}} [/tex] (3)
Let v= the speed with which the ball strikes the ground.
-v = -V₀ - gt
v = V₀ + gt
[tex]v = V_{0}- \frac{V_{0}}{2} + \frac{V_{0}}{2}\sqrt{1+ \frac{4gh}{V_{0}^{2}}} \\ v = \frac{V_{0}}{2} (1+\sqrt{1+ \frac{4gh}{V_{0}^{2}} })[/tex]
Answer:
The speed with which the ball strikes the ground is
v = V₀√[1 + (2gh)/V₀²], when the ball is thrown upward
[tex]v = \frac{V_{0}}{2} (1+\sqrt{1+ \frac{4gh}{V_{0}^{2}}}) [/tex] when the ball is thrown downward.
The difference in time of flight is
[tex]\Delta t = \frac{3V_{0}}{2g} + \frac{V_{0}}{2g} \sqrt{1 + \frac{4gh}{V_{0}^{2}} }[/tex]
All measurements are positive upward.
g = acceleration due to gravity.
Case A; The ball is thrown vertically upward.
The time, t₁, to reach maximum height is one half of the time of flight.
Because the vertical velocity is zero at maximum height,
V₀ - gt₁ = 0
t₁ = V₀/g.
The time of flight is
t = 2t₁ = (2V₀)/g. (1)
Let v = the vertical velocity with which the ball strikes the ground. Then
v² = V₀² + 2(-g)(-h)
v = √(V₀² + 2gh) (2)
Case B: The ball is thrown vertically downward.
The time of flight, t, is given by
-h = -V₀t + (-g)t²
gt² + V₀t - h = 0
t = 1/(2g)[-V₀ +/- √(V₀² + 4gh)]
Reject negative time
[tex]t=- \frac{V_{0}}{2g} + \frac{V_{0}}{2g} \sqrt{1+ \frac{4gh}{V_{0}^{2}} [/tex] (3)
Let v= the speed with which the ball strikes the ground.
-v = -V₀ - gt
v = V₀ + gt
[tex]v = V_{0}- \frac{V_{0}}{2} + \frac{V_{0}}{2}\sqrt{1+ \frac{4gh}{V_{0}^{2}}} \\ v = \frac{V_{0}}{2} (1+\sqrt{1+ \frac{4gh}{V_{0}^{2}} })[/tex]
Answer:
The speed with which the ball strikes the ground is
v = V₀√[1 + (2gh)/V₀²], when the ball is thrown upward
[tex]v = \frac{V_{0}}{2} (1+\sqrt{1+ \frac{4gh}{V_{0}^{2}}}) [/tex] when the ball is thrown downward.
The difference in time of flight is
[tex]\Delta t = \frac{3V_{0}}{2g} + \frac{V_{0}}{2g} \sqrt{1 + \frac{4gh}{V_{0}^{2}} }[/tex]
